\(M=\left(x^2-2xy+y^2\right)+\left(x^2-4x+4\right)+6=\left(x-y\right)^2+\left(x-2\right)^2+6\)

\(\min M=6\) với \(x=y=2\)

\(M=2x^2+y^2-2xy-4x-10\)

\(=x^2-2xy+y^2+x^2-4x+4-14\)

\(=\left(x-y\right)^2+\left(x-2\right)^2-14\ge-14\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}x-y=0\\ x-2=0\end{cases}\Rightarrow x=y=2\)

\(M=\left(x-y\right)^2-2\left(x-y\right)+1+9=\left(x-y-1\right)^2+9\ge9\)

\(\min M=9\)

\(M=x^2+y^2-2xy-2x+2y+10\)

\(=\left(x-y\right)^2-2\left(x-y\right)+1+9\)

\(=\left(x-y-1\right)^2+9\ge9\forall x,y\)

Dấu '=' xảy ra khi x-y-1=0

=>x=y+1

2x² + y² + 2xy - 6x - 2y + 10

= x² + y² + 1² + 2xy - 2x - 2y + x² - 4x + 4 + 5

= (x + y - 1)² + (x - 2)² + 5  5

Dấu ''='' xảy ra khi  

Vậy Min = 5 khi x = 2 và y = - 1

Ta có: \(B=2x^2+y^2-2xy+6x+10\)

\(=x^2-2xy+y^2+x^2+6x+9+1\)

\(=\left(x-y\right)^2+\left(x+3\right)^2+1\ge1\forall x\)

Dấu '=' xảy ra khi x=y=-3

Vậy: \(B_{min}=1\) khi (x,y)=(-3;-3)

\(A=2x^2+y^2-2xy-2x+y-12\)

\(A=\left(x^2-2xy+y^2\right)+x^2-2x+y-12\)

\(A=\left[\left(x-y\right)^2-2\left(x-y\right).\frac{1}{2}+\frac{1}{4}\right]+\left(x^2-x+\frac{1}{4}\right)-\frac{25}{2}\)

\(A=\left(x-y-\frac{1}{2}\right)^2+\left(x-\frac{1}{2}\right)^2-\frac{25}{2}\)

Do \(\left(x-y-\frac{1}{2}\right)^2\ge0\forall x;y\)

     \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\)

\(\Rightarrow A\ge-\frac{25}{2}\)

Dấu "=" xảy ra khi :  \(\hept{\begin{cases}x-y-\frac{1}{2}=0\\x-\frac{1}{2}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=0\end{cases}}\)

Vậy  \(A_{Min}=-\frac{25}{2}\Leftrightarrow\left(x;y\right)=\left(\frac{1}{2};0\right)\)

\(A=-2x^2-y^2-2xy-2x+y-12\)

\(-A=2x^2+y^2+2xy+2x-y+12\)

\(-A=\left(x^2+2xy+y^2\right)+x^2+2x-y+12\)

\(-A=\left[\left(x+y\right)^2-2\left(x+y\right).\frac{1}{2}+\frac{1}{4}\right]+\left(x^2+3x+\frac{9}{4}\right)+\frac{19}{2}\)

\(-A=\left(x+y-\frac{1}{2}\right)^2+\left(x+\frac{3}{2}\right)^2+\frac{19}{2}\)

Do  \(\left(x+y-\frac{1}{2}\right)^2\ge0\forall x;y\)

      \(\left(x+\frac{3}{2}\right)^2\ge0\forall x\)

\(\Rightarrow-A\ge\frac{19}{2}\Leftrightarrow A\le-\frac{19}{2}\)

Dấu "=" xảy ra khi :  \(\hept{\begin{cases}x+y-\frac{1}{2}=0\\x+\frac{3}{2}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{3}{2}\\y=2\end{cases}}\)

Vậy \(A_{Max}=-\frac{19}{2}\Leftrightarrow\left(x;y\right)=\left(-\frac{3}{2};2\right)\)

\(2x^2+y^2+2xy-6x-2y+10\)

\(=\left(x^2-4x+4\right)+\left(x^2+y^2+1+2xy-2y-2x\right)+5\)

\(=\left(x-2\right)^2+\left(x+y-1\right)^2+5\ge5\)

a) \(2x^2+y^2+4x-2y-2xy+10\)

\(=x^2+x^2+y^2+4x-2y-2xy+4+6\)

\(=\left(x^2-2xy+y^2\right)+\left(x^2+4x+4\right)-2\left(y-3\right)\)

\(=\left(x-y\right)^2+\left(x+2\right)^2-2\left(y-3\right)\)

.......................chắc không phải cách làm này đâu!

b) \(5x^2+y^2+2xy-4x\)

\(=x^2+4x^2+y^2+2xy-4x\)

\(=\left(x^2+2xy+y^2\right)+x^2-4x\)

\(\left(x+y\right)^2+x^2-4x\)

a, \(2x^2\)+\(y^2\)+\(4x-2y-2xy+10\)\(=y^2\)\(-x^2\)\(-1+2x-2y-2xy+3x^2+2x+11\)\(=\left(y-x-1^{ }\right)^2\)\(+3\left(x^2+\frac{2}{3}x+\frac{1}{9}\right)+\frac{32}{3}\)\(=\left(y-x-1\right)^2+3\left(x+\frac{1}{3}\right)^2+\frac{32}{3}\)\(\ge\frac{32}{3}\)

VẬY GTNN CỦA BIỂU THỨC \(=\frac{32}{3}\)KHI \(y-x-1=0;x+\frac{1}{3}=0\Rightarrow x=\frac{-1}{3};y=\frac{2}{3}\)

2) \(P=\frac{4}{2x^2+2xy+y^2+5x+20}=\frac{4}{\left(x^2+2xy+y^2\right)+\left(x^2+5x+\frac{25}{4}\right)+\frac{75}{4}}\)

\(=\frac{4}{\left(x+y\right)^2+\left(x+\frac{5}{2}\right)^2+\frac{75}{4}}\)

Để P đạt GTLN 

=> Mẫu thức đạt GTNN

mà \(\left(x+y\right)^2+\left(x+\frac{5}{2}\right)^2+\frac{75}{4}\ge\frac{75}{4}\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x+y=0\\x+\frac{5}{2}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{5}{2}\\y=\frac{5}{2}\end{cases}}\)

Thay x = -5/2 và y = 5/2 vào P 

Khi đó P = \(\frac{4}{\left(-\frac{5}{2}+\frac{5}{2}\right)^2+\left(-\frac{5}{2}+\frac{5}{2}\right)^2+\frac{75}{4}}=\frac{4}{\frac{75}{4}}=\frac{16}{75}\)

Vậy Max P = 16/75 <=> x = -5/2 ; y = 5/2

1) Ta có P = x² + 2xy + 3y² + 5y + 10

= (x² + 2xy + y²) + (2y² + 5y + 10) 

= \(\left(x+y\right)^2+2\left(y^2+\frac{5}{2}y+5\right)=\left(x+y\right)^2+2\left(y^2+\frac{5}{2}y+\frac{25}{16}+\frac{55}{16}\right)\)

= \(\left(x+y\right)^2+2\left(y+\frac{5}{4}\right)^2+\frac{55}{8}\ge\frac{55}{8}\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x+y=0\\y+\frac{5}{4}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{5}{4}\\y=-\frac{5}{4}\end{cases}}\)

Vạy Min P = 55/8 <=> x = 5/4 ; y = -5/4 

M=\(\left(x^2+2xy+y^2\right)+\left(2x+2y\right)+1+\left(4x^2-4x+1\right)+2014\)

=\(\left(\left(x+y\right)^2+2\left(x+y\right)+1\right)+\left(2x-1\right)^2+2014\)

=\(\left(x+y+1\right)^2+\left(2x-1\right)^2+2014\ge2014\)

\(\Rightarrow M\ge2014\Leftrightarrow minM=2014\)

\(\Leftrightarrow\hept{\begin{cases}x+y+1=0\\2x-1=0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=0,5\\y=1,5\end{cases}}\)